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Calculation of bearing loads and joint forces

This exercise addresses the following questions:

  • How do you calculate joint forces?
  • How do I calculate a frame with two fixed bearings?
  • What are the different approaches to calculating joint forces?

Task

The forces F1 to F4 act on a frame with two fixed bearings and a joint. Determine the bearing loads and the forces in joint C!

F1 = F4 = 10 kN, F2 = 30 kN, F3 = 60 kN, a = 1 m

Frame with external loads and two fixed bearings
Frame with external loads and two fixed bearings

Solution

The solution can be done in different ways, here two variants are shown:

Solution variant 1 considers both areas to the left and right of the joint separately from the beginning and sets up the corresponding equations.

Solution variant 2 sets up the equilibrium conditions for the entire system as the first step, and then releases the forces of the sub-areas.

Variant 1

The following video is in German language.

Lagerreaktionen und Gelenkkräfte berechnen - Technische Mechanik 1, Übung 22

The frame will be divided into two sections to determine the forces.

Bearing reactions and joint forces
Bearing reactions and joint forces
Balance of forces section I
x-direction

\[\tag{1} 0={F_{\mathit{Cx}}}+{F_{\mathit{Ax}}}+{F_1}\]

y-direction

\[\tag{2} 0={F_{\mathit{Cy}}}+{F_{\mathit{Ay}}}-{F_2}\]

Moments around A (left-turning moments are positive)

\[\tag{3} 0=2 {F_{\mathit{Cy}}} a-3 {F_{\mathit{Cx}}} a-{F_2} a-2 {F_1} a\]

Balance of forces section II
x-direction

\[\tag{4} 0=-{F_{\mathit{Cx}}}+{F_{\mathit{Bx}}}-{F_4}\]

y-direction

\[\tag{5} 0=-{F_{\mathit{Cy}}}+{F_{\mathit{By}}}-{F_3}\]

Moments around B (left-turning moments are positive)

\[\tag{6} 0=4 {F_{\mathit{Cy}}} a+3 {F_{\mathit{Cx}}} a+{F_4} a+2 {F_3} a\]

Solve equation %o1 for FAx

\[\tag{7} {F_{\mathit{Ax}}}=-{F_{\mathit{Cx}}}-{F_1}\]

Solve equation 3 for FCx

\[\tag{8} {F_{\mathit{Cx}}}=\frac{2 {F_{\mathit{Cy}}}-{F_2}-2 {F_1}}{3}\]

Substitute FCx in equation 7

\[\tag{9} {F_{\mathit{Ax}}}=-\frac{2 {F_{\mathit{Cy}}}-{F_2}-2 {F_1}}{3}-{F_1}\]

Solve equation 6 for FCy

\[\tag{10} {F_{\mathit{Cy}}}=-\frac{3 {F_{\mathit{Cx}}}+{F_4}+2 {F_3}}{4}\]

Substitute FCy in equation 8

\[\tag{11} {F_{\mathit{Cx}}}=\frac{-\frac{3 {F_{\mathit{Cx}}}+{F_4}+2 {F_3}}{2}-{F_2}-2 {F_1}}{3}\]

… and solve for FCx

\[\tag{12} {F_{\mathit{Cx}}}=-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\]

Solve equation 4 for FBx

\[\tag{13} {F_{\mathit{Bx}}}={F_{\mathit{Cx}}}+{F_4}\]

… and substitution of FCx yields FBx

\[\tag{14} {F_{\mathit{Bx}}}={F_4}-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\]

Substitution of FCx in equation 10

\[\tag{15} {F_{\mathit{Cy}}}=-\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_4}+2 {F_3}}{4}\]

Solve equation 5 for FBy

\[\tag{16} {F_{\mathit{By}}}={F_{\mathit{Cy}}}+{F_3}\]

… and substitution of FCy

\[\tag{17} {F_{\mathit{By}}}=\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}+{F_3}\]

F_Cy in equation 9 gives

\[\tag{18} {F_{\mathit{Ax}}}=\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}-{F_1}\]

Solve equation 2 for FAy

\[\tag{19} {F_{\mathit{Ay}}}={F_2}-{F_{\mathit{Cy}}}\]

… and substitution of FCy yields the last missing parameter

\[\tag{20} {F_{\mathit{Ay}}}={F_2}-\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}\]

Summary of the results

\[\tag{21} {F_{\mathit{Ax}}}=\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}-{F_1}\]

\[\tag{22} {F_{\mathit{Ay}}}={F_2}-\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}\]

\[\tag{23} {F_{\mathit{Bx}}}={F_4}-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\]

\[\tag{24} {F_{\mathit{By}}}=\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}+{F_3}\]

\[\tag{25} {F_{\mathit{Cx}}}=-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\]

\[\tag{26} {F_{\mathit{Cy}}}=\frac{-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{3}+{F_2}+2 {F_1}}{2}\]

Inserting the numerical values ​​delivers

\[\tag{27} {F_{\mathit{Ax}}}=15.56 \mathit{kN}\]

\[\tag{28} {F_{\mathit{Ay}}}=43.33 \mathit{kN}\]

\[\tag{29} {F_{\mathit{Bx}}}=-15.56 \mathit{kN}\]

\[\tag{30} {F_{\mathit{By}}}=46.67 \mathit{kN}\]

\[\tag{31} {F_{\mathit{Cx}}}=-25.56 \mathit{kN}\]

\[\tag{32} {F_{\mathit{Cy}}}=-13.33 \mathit{kN}\]


Variant 2

This solution is calculated here exclusively with the variables, since the numerical values ​​have already been calculated above.

The whole frame as a system with the bearing reactions

Establishing the balance of forces and moments for the entire system

\[\tag{1} \sum F_x = 0={F_{\mathit{Bx}}}+{F_{\mathit{Ax}}}-{F_4}+{F_1}\]

\[\tag{2} \sum F_y = 0={F_{\mathit{By}}}+{F_{\mathit{Ay}}}-{F_3}-{F_2}\]

\[\tag{3} \sum M(A) = 0=6 {F_{\mathit{By}}} a+{F_4} a-4 {F_3} a-{F_2} a-2 {F_1} a\]

The vertical support reaction FBy can thus be calculated directly:

\[\tag{4} {F_{\mathit{By}}}=-\frac{{F_4}-4 {F_3}-{F_2}-2 {F_1}}{6}\]

FBy is now substituted into equation 1, whereby FAy follows.

\[\tag{5} 0={F_{\mathit{Ay}}}-\frac{{F_4}-4 {F_3}-{F_2}-2 {F_1}}{6}-{F_3}-{F_2}\]

\[\tag{6} {F_{\mathit{Ay}}}=\frac{{F_4}+2 {F_3}+5 {F_2}-2 {F_1}}{6}\]

Bearing reactions and joint forces
Investigation of the sections I and II

Next, the balance of forces in the y-direction is set up for section I in order to calculate the joint force FCy:

\[\tag{7} \sum F_y = 0={F_{\mathit{Cy}}}+{F_{\mathit{Ay}}}-{F_2}\]

\[\tag{8} {F_{\mathit{Cy}}}=-\frac{{F_4}+2 {F_3}-{F_2}-2 {F_1}}{6}\]

The sum of moments around A leads to FCx.

\[\tag{9} \sum M(A) = 0=2 {F_{\mathit{Cy}}} a-3 {F_{\mathit{Cx}}} a-{F_2} a-2 {F_1} a\]

\[\tag{10} {F_{\mathit{Cx}}}=-\frac{{F_4}+2 {F_3}+2 {F_2}+4 {F_1}}{9}\]

The equilibrium of forces in the x-direction for section I results in the horizontal support reaction in A.:

\[\tag{11} 0={F_{\mathit{Cx}}}+{F_{\mathit{Ax}}}+{F_1}\]

\[\tag{12} {F_{\mathit{Ax}}}=\frac{{F_4}+2 {F_3}+2 {F_2}-5 {F_1}}{9}\]

FAx is now substituted into equation 2, whereby FBx follows.

\[\tag{13} {F_{\mathit{Bx}}}=\frac{8 {F_4}-2 {F_3}-2 {F_2}-4 {F_1}}{9}\]

The second solution leads to the desired results faster!